Career Guide Advanced Finance and Quant Finance Interviews

Customizedfor:Benjamin(ybl205@nyu.edu)
© 2002 Vault Inc.
ADVA
FINAN
QUAN
VAULT GUIDE TO
ADVANCED
FINANCE AND
QUANTITATIVE
INTERVIEWS

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Printed in the United States of America

INTRODUCTION 1
Your First Step . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1
Problem Solving Strategies . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2
Sample Questions and Answers . . . . . . . . . . . . . . . . . . . . . . . . . .3
BOND FUNDAMENTALS 5
Bond Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7
Time Value of Money . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .9
Bond Prices and Relationships to Yields . . . . . . . . . . . . . . . . . . .15
Taylor Series Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .18
Bond Price Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .21
Sample Questions and Answers . . . . . . . . . . . . . . . . . . . . . . . . .30
Summary of Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .33
STATISTICS 35
Random Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .37
Key Statistical Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .41
Permutations and Combinations . . . . . . . . . . . . . . . . . . . . . . . . .50
Functions of Random Variables . . . . . . . . . . . . . . . . . . . . . . . . .52
Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .56
Regression Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .59
Sample Questions and Answers . . . . . . . . . . . . . . . . . . . . . . . . .72
Summary of Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .78
Vault Guide to Advanced Finance and Quantitative Interviews
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C A R E E R
L I B R A R Y

DERIVATIVES 81
Introduction to Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . .83
Forward Contracts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .83
Futures Contracts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .89
Swaps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .95
Options: An Overview+RC . . . . . . . . . . . . . . . . . . . . . . . . . . . .96
Combining Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .110
Option Valuation I: Introducing Black-Scholes . . . . . . . . . . . . .121
Option Valuation II: Other Solution Techniques . . . . . . . . . . . .130
Option Sensitivities: the Greeks . . . . . . . . . . . . . . . . . . . . . . . .155
Exchange-traded options . . . . . . . . . . . . . . . . . . . . . . . . . . . . .160
Exotic Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .166
Sample Questions and Answers . . . . . . . . . . . . . . . . . . . . . . . .170
Summary of Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .180
FIXED INCOME 187
Bond and Fixed Income Market Issuers . . . . . . . . . . . . . . . . . .189
Types of Fixed Income Securities . . . . . . . . . . . . . . . . . . . . . . .195
Quoting Bond and Fixed Income Prices . . . . . . . . . . . . . . . . . .196
Analyzing and Valuing Bonds and Fixed Income Securities . . .196
Fixed-income Specific Derivatives . . . . . . . . . . . . . . . . . . . . . .205
Mortage-backed Securities . . . . . . . . . . . . . . . . . . . . . . . . . . . .229
Summary of Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .248
Equity Markets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .249
Equity Valuation Overview . . . . . . . . . . . . . . . . . . . . . . . . . . .251
Dividend Discount Model . . . . . . . . . . . . . . . . . . . . . . . . . . . .253
Multiples Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .254
Return on an Asset . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .258
CAPM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .259
Equity Indexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .266
Hybrids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .268
Equity-specific derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . .273
Table of Contents
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C A R E E R
L I B R A R Y

CURRENCY AND COMMODITY MARKETS 283
Currency Swaps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .285
Comodity Swaps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .288
RISK MANAGEMENT 297
Measuring Market Risk: Value at Risk . . . . . . . . . . . . . . . . . . .299
Types of Risk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .302
Currency Risk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .304
Fixed-income Risk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .305
Equity Risk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .310
Currency and Commodity Risk in the News . . . . . . . . . . . . . . .313
Hedging Risk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .317
Portfolio Risk and Correlation . . . . . . . . . . . . . . . . . . . . . . . . .320
Arbitrage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .322
Sample Questions and A+R[-47]Cnswers . . . . . . . . . . . . . . . . .324
APPENDIX
Advanced Finance Glossary
About the Author
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C A R E E R
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INTRODUCTION

Vault Guide to Advanced and Quantitative Finance Interviews
Introduction
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1
Your First Step
Congratulations on taking your first step to succeeding in your advanced finance interviews. This book
was written to give you the technical background needed to master that interview – compiled into one
convenient volume. Quantitative and Wall Street interviews are notoriously tough, and with good reason.
These types of jobs pay very well – and a lot of people want them.
This book will give you the edge you need to succeed. This is the book the writers and editors wish they
had when they were interviewing. It is the distillation of years of experience in the finance field, in
teaching finance and in numerous interviews. Vault editors have even taken interviews just to find out
what kinds of questions interviewers are currently asking, in order to bring you the latest in this book.
In quantitative interviews, mastery of the subject matter is assumed – it is your starting point. You will also
have to convince your interviewer(s) you are the right fit for the firm and have the experience and
background that they are looking for. Of course, no book can give you that – though the Vault Guide to
Finance Interviews gives you helpful pointers in that direction.
What this book can do is help you review and master the required subject matter, without which no amount
of charm will get you by. (Although charm is always good.)
Also unique to this book are strategies to help you succeed on those tough interview questions that you may
not be prepared for. Some questions you may get are deliberately designed to be impossible to answer.
The interviewer just wants to see how you think and how you approach problems. Remember, all of the
easy problems have already been solved. The problems you will see on the job will likely be things that no
one has quite seen before.
Still, you will find some interviewers who will ask questions straight out of textbooks (one insider reports
receiving the following question in a recent interview with Bloomberg: “What is an equivalence statement
in FORTRAN and why would it be used?”) It is simply the style of certain companies and interviewers to
ask questions from textbooks, so you should be prepared for this if you want to land a job. For inside
information on interviewer style, you may want to check out the Vault message boards. For everything
else, let this book be your guide. Wherever possible, we’ve used questions from actual interview
experience, including the interviewer’s comments on what they were looking for (when we could get it).
It is our hope that you will find the problem solving strategies and the material in this book indispensable to
you even after you land your job. Good luck!

Introduction
2
Problem Solving Strategies
What do you do when confronted with an interview question you have absolutely no idea how to solve?
We recommend the following strategies – it should help you handle most anything thrown at you.
Strategy #1 Cite from memory
Strategy #2 Draw a figure
Strategy #3 Work backwards
Strategy #4 Formulate an equivalent problem
Strategy #5 Enumerate all cases
Strategy #6 Search for a pattern
Strategy #7 Bracket the answer – solve the extreme cases
Strategy #8 Relate to something you know
Strategy #9 Take advantage of symmetry
Remember to RELAX. Try to see these interviews simply as conversations. It is a chance for interviewers
to evaluate you, but remember, you are also deciding if you want to work there as well. The more relaxed
and calm you are, the easier it will be for you to think creatively, which is often what is required in finance
interviews. Also, try to think of the tough interview questions as amusing little problems (the interviewer
probably does). One recent interviewee reports having an interviewer grill her relentlessly on currency
forwards, interest rate parity and so on. When the job seeker finally reported being unsure of the approach
to one question, “The interviewer laughed and said, ‘Don’t worry. If you had known the answer to this
problem, I would have found something else that you don’t know. That’s my job.’”
Remember, sometimes you will be able to use one strategy by itself to answer a question, but imagine what
a powerful approach it is when you can combine two or more. Those tough interview questions won’t have
a chance. You will see the above strategies used throughout this book, and identified to help you remember
them. Often, problems can be approached from more than one angle, so don’t feel that you must use the
approach we show.

Introduction
3
Sample Questions
1. You have a sheet of paper and an infinite supply of tokens. I also have an infinite supply of
tokens. We take turns placing tokens on the paper, one token at a time. We cannot place tokens on
top of other tokens (no overlapping), and the tokens cannot extend over the edges of the paper. The
last player to place a token on the paper wins. What is your winning strategy? (This is called a
“strategy game” question, and is an actual question recently asked on a hedge fund interview.)
Solution: Don’t freak out if you see something like this. The interviewer is just trying to get a sense of
how you attack a new problem. Let’s go through our list of tactics. Tactic #1 will not work here. Tactic
#2 has promise: Try breaking it down into smaller sub-problems. What if the paper were so small that
only a single coin could fit on it?
In this case your strategy would be to go first. After you place your coin, your opponent has no place to
place his, and you win. Next, what if the paper were big enough for two coins? Here, you place your coin
in the dead center of the sheet so your opponent can’t place his coin. Again, your strategy would be to go
first.
This tactic can be repeated until you have derived the correct answer: You always move first, and if you
play the game properly, you will always win.
2. What do you think is the major factor impacting the profitability of an airline? (This was an actual
question asked in a Goldman Sachs equity quantitative research interview.)
Solution: This is another question that the interviewer doesn’t expect you to have memorized, but expects
you to go through a reasoning process enumerating possible factors affecting airline profitability to come
up with the most important one. You could say, “passenger meals, labor costs, weather delays, leasing
costs, marketing, maintenance, price wars,” but the major cost driver is probably “fuel.”
3. Would the volatility of an enterprise be higher or lower than the volatility of its equity? (Actually
asked by a Goldman Sachs interviewer who kept coming back to this in one form or another during the
interview.)
Prohibited

Introduction
4
Solution: This is a straightforward Statistics or Corp Finance 101 question. Even if you have never seen
this exact question before, it can be reasoned out. In the following response we employ a combination of
tactics #1 and #5.
Corporations usually have both debt and equity (we reason.) So, suppose you have a portfolio consisting of
w percent equity and (1-w) percent debt. We calculate volatility as the square of the standard deviation of
stock returns. Then
( ) ( ) DEEDDEp Covwwww sssss ,
222
121 -+-+=
Assume that the covariance is zero. Then ( ) 222
1 DEp ww sss -+=
Now, consider three cases.
Case one: There is no debt.
Then the variance of the enterprise (sp above) is equal to the variance of the equity.
Case two: There is no equity.
Then the variance of the enterprise (sp above) is equal to the variance of the debt.
Case three: There is a combination of the two.
We have bracketed the answer already in cases one and two: the result must lie in between these. Now a
judgment must be made. Assume that the volatility of the debt is lower than the volatility of the equity.
Then the volatility of the enterprise with both debt and equity will be lower than the volatility of the equity
alone, since the volatility of the enterprise is somewhat of a weighted average of both debt and equity.
Also, note that we multiply the volatility of the equity by w, a fraction assumed to be less than one. The
only way that we could have volatility of the enterprise higher than that of equity alone would be if sD > sE
and if w were negative (impossible). To see this, rearrange the equation to ( ) 2222
DDEp w ssss +-= . If the
vol of debt equals vol of equity, vol of the enterprise still equals vol of the equity.

ADVA
FINAN
QUAN
BOND
FUNDAMENTALS

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Bond Fundamentals
7
This is the chapter that you will need to know if an interviewer or headhunter asks, “Do you know bond
math?” – and you want to answer, “Yes.”
Bond Basics
A bond is a contract to provide payments according to a specific schedule. Bonds are long-term securities
with maturities exceeding one year, in contrast to bills, or other short-term debt such as commercial paper,
which have maturities of less than one year. The bond universe is huge. There are treasury bonds, agency
bonds, junk bonds, corporate bonds, zero coupon bonds, municipal bonds, sovereign bonds, tax-free bonds
and so forth. In addition to all of these, there are options on bonds, options on options on bonds, and so on.
(These will be covered in detail in later sections.) Finally, most bonds are highly sensitive to interest rates,
so we will have to study the yield curve in some detail. For now, we will cover the fundamental financial
concepts required in valuation of bonds.
Bonds are different from equity
In the contractual agreement of a bond, there is a stated maturity and a stated par value. This is unlike an
equity, which has no maturity and no guaranteed price at maturity. To express this definite price at
maturity, we say that bonds converge to par value at maturity. This defined par value makes the volatility
of a bond generally lower than a share of stock (equity), especially as maturity draws close.
However, don’t get the idea that bonds are without risk or uninteresting. Quite the contrary. According to
a February 28, 2000, BusinessWeek Online article (“Is the Bond Market Ready for Day Traders?”), “Bonds
are no longer the stodgy investments they once were… What most people don't know is that the 30-year
Treasury bond has the same volatility as an Internet stock.” Constructing models for bond valuation is one
of the tougher challenges out there. Bonds have many inherent risks, including default risk, basis risk,
credit risk, interest rate risk and yield curve risk, all of which may not apply to equity or equity-like
securities.
Bond ratings
Bonds are generally considered to be less risky than equity (except for junk bonds), so they can generally
be expected to have lower rates of return. In the world of bonds, we are concerned with the credit rating of
the company or municipality that issued the bond. Credit ratings are provided by major ratings agencies,
including Moody’s, Standard & Poor’s, and Fitch. You may wish to familiarize yourself with these ratings
(http://www.standardandpoors.com/, http://www.moodys.com/, http://www.fitch.com.
Recently, a Goldman Sachs interviewer quizzed one of this book’s editors on ratings of corporate bonds
and subsidiary liability in case of default. Of course, you may not have to worry about this if you do not
have this type of experience listed on your resume, but remember, anything on your resume, no matter how
long ago or obscure, is fair game.)
Bond ratings affect the ease and cost of obtaining credit for the corporation issuing the bonds. The higher
the rating – AAA is the highest S&P rating, for example – the lower the cost of credit. As the corporation’s
credit rating declines, it gets more and more expensive for the corporation to raise new funds. Most
corporate treasuries are concerned with possible ratings downgrades and check frequently with ratings
agencies before undertaking something that could potentially result in a downgrade. Downgrades can also
affect investors, as many fixed income managers in asset management firms have mandates to hold only
corporate-grade bonds and better. If a corporation’s bonds fall to the “junk” category (see, for example,
Xerox, May 2002), the institutional investors in the company may have to sell their holdings to comply

Bond Fundamentals
8
with client requirements. This dumping – which could be large holdings of the bonds – makes the price of
the bonds drop.
Before pushing forward with valuation of bonds under scenarios such as the above, we have to review the
time value of money, which is possibly the most important concept in finance. You will see this over and
over again in various forms, so there’s no time like the present.

Bond Fundamentals
9
Time Value of Money
Discounting, present and future value
If you deposit $1,000 in a bank or money market account today, you expect to earn some interest on your
investment so that, as time passes, the value of your investment grows. (If it did not, you would probably
not put your money in the bank.) If you are earning a rate of 10% a year, at the end of the year you will
have your original $1,000 plus the interest earned, 10% of the principal invested ($1,000) or $100. The
total value of your investment is then $1,000 + $100 = $1,100 = $1,000(1 + r), where r is the interest rate in
percent for the period. In general, the future value of an initial investment P0 over n compounding periods
is given by the formula
where:
n = number of compounding periods in time interval t
t = time interval
r = the interest rate earned per compounding period (assumed constant)
P0 = the initial value (principal)
PN = the final value
The above formula then gives the future value of money. This type of formula is called simple
compounding. Many important results in finance are based on this very simple principal of the time value
of money.
NOTE: The value PN is often called “the future value” and P0 “the present value”. So we could also write
Example (Annual Compounding) What is the value of $1,000 after one year if interest is only compounded
once per year? Here n = 1, t = 1 year, r = 10%/year, P0 = $1,000.
Example (Semi-Annual Compounding): What is the value of $1,000 after one year if interest is
compounded twice per year? Here n = 2 and:
Example (Quarterly Compounding) What is the value of $1,000 after one year if interest is compounded
four times per year? Here n = 4 and:
nt
n
r
PVFV ÷
ø
ö
ç
è
æ
+= 1
1
1
1
10.0
1000,1$ ÷
ø
ö
ç
è
æ
+=P =$1,100
2
1
2
10.0
1000,1$ ÷
ø
ö
ç
è
æ
+=P =$1,102.5
4
1
4
10.0
1000,1$ ÷
ø
ö
ç
è
æ
+=P =$1,103.81
nt
N
n
r
PP ÷
ø
ö
ç
è
æ
+= 10

Bond Fundamentals
10
Example (Daily Compounding) What is the value of $1,000 after one year if interest is compounded each
trading day? Here n = 250 and:
The value of the investment increases as we compound more and more frequently since interest is being
compounded on interest. In the limit as n approaches infinity, we have continuous compounding, which
gives the future value of money as:
Example (Continuous Compounding) What is the value of $1,000 after one year if interest is compounded
continuously?
1*1.0
1 000,1$ eP = =$1,105.17. It should not surprise you that this answer is very close to the result that we
obtained with daily compounding.
What if you will be investing over a period of time in, say a savings account for retirement? To develop
the formula, suppose you invest $1,000 for the next five years at a constant 10%/year. How much will you
have at the end of the year? Assume you invest at the beginning of each year and interest is paid once per
year.
Year Zero: Initial $1,000 invested, will stay in account for five years.
Year One: Another $1,000 invested, will stay in account for four years.
Year Two: Another $1,000 invested, will stay in account for three years.
Year Three: Another $1,000 invested, will stay in account for two years.
Year Four: Another $1,000 invested, will stay in account for one year.
So our account will contain an amount of
$1,000(1+0.1)5
+$1,000(1+0.1)4
+$1,000(1+0.1)3
+$1,000(1+0.1)2
+$1,000(1+0.1)1
=$6,715.6 after five years.
In terms of a formula,
since the investments at each year (CFi) are the same each year.
This is an interesting application for retirement saving: If a 20-year old could earn 10%/year on average,
how much would they have at age 65?
Solve the problem by figuring how much there would be after 20 years, then use the simple compounding
formula to take it forward another 25 years.
What if a person waits until age 40 to start saving for retirement? How much would they have to save per
year to end up with the same amount as the saver who began at age 20?
Solving this, you’ll understand why so many people are planning to work past age 65 these days.
Present value of a future dollar The same formulas can be used to solve for the present value of a future
payment. We just have to solve for P0. For n compounding intervals per year,
250
1
250
10.0
1000,1$ ÷
ø
ö
ç
è
æ
+=P =$1,105.15
rt
t ePP 0=
( ) ( )
in
i
i
n
i
it rCFrCFP åå ==
+=+=
11
11

Bond Fundamentals
11
and for continuous compounding, P0 = Pte-rt
.
NOTE: In this case, the rate r is called the discount rate since P0 < Pt always.
Example You are promised a payout of $1,000,000 ten years from now. (Financial application: this is
used to value a zero coupon bond.) If the discount rate is 10%, what is this payout worth today?
Use the continuously compounded formula. 10*1.0
0 000,000,1$ -
= eP =$367,879. You should be indifferent
between a payout today of this amount and a future payment of $1,000,000 in ten years.
Now, what if you have a series of cash payments? (Either these terminate at some future time or go on to
infinity -- such types of payments are called perpetuities.)
In the first case, suppose you buy a bond paying 8% per year for the next 10 years. At the end of 10 years,
you will receive your last interest payment plus return of your principal. Assume that the principal is
$1,000. What is this bond worth today? (Later we will see that there are three scenarios that can occur
depending on what the investment rate r is. For now, assume that you can earn 10% by placing money in a
savings bank.) All we do is take each payment and discount it back to the present. We are paid
0.08*$1,000 each year or $80 ( the “coupon payment”.) We discount the first payment over a one-year
period, the second payment over a two-year period and so on. At the end of 10 years, we have $80 plus
the return of our principal for a total of $1,080 to be discounted back ten years. It is really like working ten
independent problems and summing together.
where the values of CFi for i = 1 to n-1 are the coupon payments, and the last cash flow, CFn, is that year’s
coupon payment plus the return of par. Observe that this formula reduces to the one we had earlier when n
=1. Also we have assumed that the discount rate r is constant over the life of the investment.
Technical Note: To be more general, we should actually discount each year by the prevailing discount rate,
ri, at year i. Then we have:
The preceding formula is very important and will be used over and over.
Annuity: If the payments CF are constant over a period, this is called an annuity. Common examples
include mortgage and car loan payments.
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )10987654321
1
080,1$
1
80$
1
80$
1
80$
1
80$
1
80$
1
80$
1
80$
1
80$
1
80$
rrrrrrrrrr
PV
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
=
PV = $877. In terms of a formula, we have
( )
å =
+
=
n
i i
i
r
CF
PV 1
1
( )
å =
+
=
n
i i
i
i
r
CF
PV 1
1
nt
N
n
r
PP
-
÷
ø
ö
ç
è
æ
+= 10

Bond Fundamentals
12
Perpetuity: Now, what if we receive a payment of CF forever? We take the limit as n approaches infinity
and use this result instead. To use this approach, r has to be constant. It turns out that
Want to check without using any calculus? Just use a spreadsheet with any i and r you want. Let n
increase until the answer stops changing (in Excel, the function PV is used as “=PV(r,n,pmt)” or
“=PV(0.1,10,80)” for r = 0.1, n = 10 and PMT = 80). You will find that you approach the value of 800, or
CF/r = $80/0.1. Remember the above formula because you will it again.
This formula can be used to value perpetual debt that a corporation may have. The corporation may have
fixed income liabilities on its books that have a finite expiration period, but if it can keep “rolling over” the
debt, it can be valued as a perpetuity. Here, the rate r is the average coupon payment on the corporation’s
debt. This formula can also be used to value a company in the mature stage where it is stable and paying
out a constant dividend. Then, r plays the role of the risk of the company, or hurdle rate. This model can
be used if it is assumed that the company is a going concern, i.e. it will operate into perpetuity.
As an example, what if a company is paying out dividends of $1.35/share at a hurdle rate of 20%? What is
the company worth on the basis of this model?
Value = $1.35/0.20 =$6.75/share.
Gordon growth model
This is used for valuing cash flows such as debt or stock dividends that are projected to grow at a constant
rate g. Then,
In the limit as n approaches infinity, we get
For example, in the above, suppose that the company’s dividend policy is to grow the dividend 10%/year
for perpetuity. The value of the company should be calculated using the Gordon Growth Model. We need
Slick Trick: Now that we know the formulas for annuities and perpetuities, we can come up with a shortcut
for calculating the value of an annuity that pays from i = 1 to n. This means we won’t have to sum that
long series again. First, look at the cash flow diagram of a perpetuity paying cash flow C at discount rate r:
( ) r
CF
r
CFn
i in
=
+
å =¾®¾¥
1
1
lim so
r
CF
PVPerpetuity =
gr
CF
PV
-
= 0
to know the current dividend, CF0 = $1.65. Then,
1.02.0
65.1$
-
=PV =$16.5/share. This should be higher than
the value with no growth, and it is.
( )
( )
( )
( )
( )
( )
( )
( )
å
= +
+
=+
+
+
+
+
+
+
+
+
=
n
i
n
n
r
g
CF
r
gCF
r
gCF
r
gCF
PV
1
03
3
0
2
2
0
1
0
1
1
1
1
1
1
1
1
L

Bond Fundamentals
13
C C C C C C C
i=0 i=1 i=2 … i=n i=n+1 … i→∞
The present value of this perpetuity (call it PV1) has already been shown to be
Next, consider a perpetuity paying C that doesn’t start until time i = n+1. The value of this
perpetuity at time n is
r
C
, and to get the present value at i=0, just discount back by dividing by
(1+r)n
. So the value of the perpetuity at i=0 (call it PV2) is
Now we can find the value of the annuity running from i = 0 to i=n. We just take the infinite
perpetuity running from time i=0 to infinity, PV1, and chop off the part we don’t want: the value of
the perpetuity running from i = n+1 to infinity. Hence,
This can be used to value a stream of coupons from a bond. The formula can be easily modified to
handle the full coupon bond by just adding on a term representing the PV of the principal
repayment, so we would have:
Example: Let’s go back to that 10-year, 8% coupon bond at a discount rate of 10%. Tedious calculations
gave its value as $877. Using the above formula gives:
=$491.56+$385.54 = $877.1
r
C
PV =1
( ) r
C
r
PV n
+
=
1
1
2
( ) ( ) ÷
÷
ø
ö
ç
ç
è
æ
+
-=
+
-= nn
rr
C
r
C
rr
C
PV
1
1
1
1
1
( ) ( )nncoupbond
r
P
rr
C
PV
+
+
÷
÷
ø
ö
ç
ç
è
æ
+
-=
11
1
1
( ) ( )1010
1.01
000,1
1.01
1
1
1.0
80
+
+
÷
÷
ø
ö
ç
ç
è
æ
+
-=coupbondPV

Bond Fundamentals
14
Bond Prices and Relationships to Yields
We have laid all of the necessary foundation for pricing bonds. In fact, we have already started pricing
bonds. It is now time to talk about three types of bonds: discount bonds, par bonds and premium bonds.
There is only one thing that differentiates these three types of bonds: the spread between the discount rate r
and the coupon rate i.
Discount bond
In the previous example, our bond sold at less than the face value of $1,000. This is an example of a
discount bond. The reason it sold for less than its face value (“less than par”) is because the coupon interest
rate, 8%, is lower than the discount rate r. Think about it: The discount rate r is the expected interest rate
an investor could earn by investing in a vehicle such as high-yield treasuries. If the investor could earn
10% elsewhere, but this bond is only paying 8%, shouldn’t the investor be compensated for taking a lower
interest rate? The bond is said to be selling at a discount to par.
Par bond
If the bond is selling for par, that is, the present value is equal to the face value, here $1,000, the bond is
said to be selling at par or a par bond. We look at the general formula
Since we know that the PV of a payment P at time n is just
( )n
r
P
+1
, the last term above, the only way that
this can be is if i = r. If this occurs, the bond is paying interest at the rate r and we have a par bond.
Premium bond
On the other hand, if the bond pays a rate that is higher than the prevailing rate r, it will be priced higher
than par and is called a premium bond.
We also have to now interject reality into the story. Unless the yield curve is flat, interest rates do change
with time to maturity (“tenor”). Let’s take a look at a yield curve from May 3, 2002. The following data is
from Bloomberg.com (but you can see yield curves in many places).
( ) ( )nn
rr
C
rr
C
PV
+
+
+
-=
1
1
1
1
= 1 if C = r where C is the coupon rate and r is the discount rate.

Bond Fundamentals
15
Tenor,
months Yield, %
3 1.75
6 1.89
24 3.21
60 4.4
120 5.09
360 5.59
Yield Curve
03 May 2002
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
0 30 60 90 120 150 180 210 240 270 300 330 360
Tenor, Months
Yield,%
The fact that the yield curve is not flat means that we have to use different discount rates to value each cash
flow. For example, if we had a cash flow at two years, we would use the rate of 3.21% from the table
above; a cash flow ten years out would be discounted at a rate of 5.09% and so on. “Missing” data in
between the given points (such as three years, four years and so forth) must be calculated. We will have
much more to say about these topics in the Fixed Income Section. For now, to make things easy, assume
that the following has been calculated:
Tenor,
months
Yield,
%
12 2.33
24 3.21
36 3.61
48 4.00
60 4.4
Let’s value a five-year bond paying a coupon of 3.5%. On a bond with face value of $1,000 we will then
receive $35 each year (assuming for simplicity that we have annual compounding.) Here is a table of our
cash flows, discount rate and present value of each cash flow.
Tenor,
months
Yield,
%
Cash
Flow
PV Cash
Flow

Bond Fundamentals
16
12 2.33 35 34.20
24 3.21 35 32.86
36 3.61 35 31.47
48 4.00 35 29.91
60 4.4 1035 834.52
PV Bond $ 962.97
Why is the bond priced below par? We have different discount rates. In some sense, the “average” rate
used must be lower than the coupon rate. We can determine this average rate by setting the present value
of the price of our bond equal to the present value of a bond with the same cash flows but a uniform rate r.
We solve by y by trial and error, using a spreadsheet and the Solver function, or with our formula for an
annuity. We find that the value of y that satisfies the above equality is 4.34%. This special y is called the
yield of the bond. That’s all yield is: just a mathematical concept that is used to allow us to compare
different bonds on a level playing field. Otherwise, how would we rank bonds? Is it correct to say that a
bond with a higher coupon is a better investment? You can’t just rank by coupon since different bonds
have different maturities. Note that on Bloomberg’s page, they show the current yield as 4.35%, very close
to ours, but we assume that they use slightly different interpolation methods.
Now we are in a position to explore the very critical relationship of price to yield.
Price and yield are inversely proportional: as yield increases, price decreases. As yield decreases,
price increases.
(We repeat this statement because it is so important. You can think of it as the “first law of bond dynamics”
if you want.)
You can see from the equation above that if yield is higher than 4.34% required to maintain the equality,
we will be dividing by a larger number so the PV (price of bond) should decrease. And if we decrease the
yield we are dividing by a smaller number, so the price will increase. Practically, this makes sense.
Bonds must converge to par at maturity
Yield is the mathematical mechanism by which we get from the present to the future. So, if a price is low,
we have to have a whopping large yield to climb to par. If price is already high, say, close to par, we don’t
need very much growth to get to par. This is such a crucial fact that you may want to build your own bond
model on a spreadsheet and explore the effects of varying yield.
Thus, since
( ) ( )n
n
i
i
y
Par
y
C
PV
+
+
+
= å
= 111
, for our example,
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )54325432
1
035,1$
1
35$
1
35$
1
35$
1
35$
044.1
035,1$
04.1
35$
0361.1
35$
0321.1
35$
0233.1
35$
97.962$
yyyyy +
+
+
+
+
+
+
+
+
=++++=

Bond Fundamentals
17
Taylor Series Expansion
Suppose you have information about a function at one point and want information about that function at
some other point. For a simple example, suppose Gordon is now located 200 miles east of Chicago. He is
traveling due west at 60 miles/hour and is very anxious to make his class. How would he be from Chicago
after one hour of driving if he decides to accelerate his speed at a steady 10 miles per hour over that hour
(so that his speed after one hour is 70 mph)? Or, consider another example: suppose you have the price of
a bond at a certain yield. What would the price of that bond be if the yield changes by one percent? You
may not realize it, but to solve these and similar problems you use the principles of Taylor Series
expansions. Taylor Series are even used to derive the Black-Scholes equation and Ito’s Lemma, which we
will come to later. In fact, if you have taken physics and are familiar with the equation of time position of a
particle x(t) = x0 + v0t + 1/2 a0 t2
, you have are already used Taylor Series.
The distance x-k must be small and the derivatives must exist at k. Note that the Taylor Series
includes an infinite number of terms. In practice, we can only take a finite number of terms, and
there will be truncation error due to the contribution of the terms that are dropped. So, f(x) is
approximated by a finite-number-of-terms Taylor Series, plus a truncation error. For example, the
second-order Taylor Series expanded about the point k is given as:
where R3 is the truncation error, consisting of the sum of terms n = 3 to infinity. Also note that if we are
just approximating polynomials of degree n, the Taylor Series of order n will give an exact result. (The
Taylor Series of order n is a polynomial of order n.) Graphically, what we are trying to do is this:
So, here is the theory that you need to know:
A continuous, differentiable function f may be expanded in a Taylor Series about a point k as follows:
( )( )( ) ( ) ( )( ) ( )( ) L+-+-+-+=-= å
¥
=
3
'''
2
''
0 !3!2
)(')(
!
)( kx
kf
kx
kf
kxkfkfkx
n
kf
xf n
n
n
( ) 3
2
3
0
)(
))((''
!2
1
))((')(
!
)(
)( RkxkfkxkfkfRkx
n
kf
xf
n
n
n
+-+-+=+-= å
¥
=

Bond Fundamentals
18
f(x)=?
k
f(k)
x
f
The assumptions are that the function exists over the range of interest of expansion (that is, it is continuous
between k, the known point, and x, the point you are trying to forecast), and that the derivatives above
exist. Note that the right hand side is completely known so you can just add it up to get your forecast for f
at the desired point. So, a Taylor Series expansion is simply a technique to make an approximation of the
behavior of the function f(x) over the interval (k,x). If you knew the actual function f(x), you could simply
evaluate it at the desired value x. The assumption is that you do not know what f will be at x, and need
some way to estimate it.
Actual practice tip: You don’t even really need to know the actual function f(x) as long as you know, or
can estimate, the values of f and its derivatives at the point k. You’ll see this in pricing bonds using
duration and convexity.
Let’s try an example. (We’ll do a math problem first, for confidence, and then we’ll move on to finance.)
Example Let the function be f(x) = x3
– x – 1. We already know that f(2) = 23
–2 – 1 = 8 – 3 = 5. To
check Taylor Series, assume that you only have information about f and its derivatives at the point x = 1
(this will be the “k” in the Taylor Series equation, the known point), and we seek the value of the function f
at 2, the unknown point.) We need to have all of the values of f and derivatives at the known point to
forecast what f will be at 2. How many derivatives is enough? The more you use, the better the
approximation, and there are formulas that tell how far you must go to fall within an acceptable error. In
this case, we will compare approximations with using the first derivative only, the first and second
derivatives, and the first three derivatives (which will give the exact solution for this cubic function).
f(x) = x3
– x – 1 f(1) = 13
– 1 – 1 = - 1
f’(x) = 3 x2
– 1 f’(1) = 3(1)2
– 1 = 2
f’(x) =6 x f’’(x) =6(1) = 6
f’’’(x) = 6 f’’’(1) = 6
Using first derivative only: f(2) ≈ f(1) + f’(1)(2-1) = -1+2(2-1) = 1. Error = 5 – 1 = 4
Using first two derivatives: f(2) ≈ f(1) + f’(1)(2-1) + f’’(1)(2-1)2
/2 = 1 + 6/2= 4 . Error = 5 – 4 = 1.
Using first three derivatives: f(2) ≈ f(1) + f’(1)(2-1) + f’’(1)(2-1)2
/2 + f’’’(1)(2-1)3
/6 = 4+6/6 = 5.
Error = 5 – 5 = 0.

Bond Fundamentals
19
Note how the error between the true (known) value and our approximation decreases as we increase the
number of derivatives used. What we are doing is providing more and more information in order to more
closely approximate the unknown value.
Also, note that it really isn’t necessary to know the definition of the function. In the real world, we might
just know the derivative values. We could have just as easily solved this just knowing the values f(1), f’(1),
f’’(1) and so on, but the values would have to be known to us in some way.
Note: If the function f depends on two variables, such as x and y, we just differentiate with respect to each.
We have
This is called the Multivariate Taylor Series. It can be used to estimate to the value of f at a point (dx,
dy) away from a point (x,y) at which f is known. It requires knowledge of the above derivatives at (x,y).
Example Let f(x,y) = x2
y. Suppose we only know the value of f for x = 1 and y = 3, so that f(1,3) = 3.
Let’s say f represents the price of an option, x the price of a stock and 1 represents the time parameter. We
are interested in estimating the value of f if x and y change a small amount, say by 1 each. (Of course, we
could just plug in the x and y values to get f(2,4) = 16, but we want to try out our Taylor Series
approximation here. Later we will use just such a method to derive the Black-Scholes equation for pricing
options.) To estimate using Taylor Series, we need the derivatives.
=6+1+3+2 = 12. This is the change in f caused by moving from (1,3) to (2,4), so the new value of f is the
sum of the old value of f plus the change, f(2,4) = f(1,3) + df = 3 +12 = 15. Note that since we truncated
the series after the second order terms, we still have truncation error to account for. But it will do for a first
approximation.
Example The price change of a bond caused by the change in yield can be estimated by expanding P as a
function of y in a Taylor Series. Using just the first two terms, we have:
What is the price change if Dy = 1%, the first derivative of P with respect to y is equal to –6,721 and the
second derivative of P with respect to y is 60,600? Just plug into the formula to get dP = -$64.18.
In following sections, we will see more Taylor Series, including finding out where the above derivatives
came from and what they mean, and in deriving the Black-Scholes equation and numerical approximations
for its solution.
Why do you need to know the Taylor Series? It is often the case that we have information about something
at a certain point, say, in time, for example, and want to know what it might be another other point in time
(this is called forecasting.) If you know the rate at which the function is changing, and have reasonable
0,2,2,,2 2
22
2
2
2
=÷÷
ø
ö
çç
è
æ
¶
¶
¶
¶
=
¶
¶
=÷
ø
ö
ç
è
æ
¶
¶
¶
¶
=÷÷
ø
ö
çç
è
æ
¶
¶
¶
¶
=
¶
¶
=÷
ø
ö
ç
è
æ
¶
¶
¶
¶
=
¶
¶
=
¶
¶
=
¶
¶
y
f
yy
f
x
x
f
yy
f
xxdy
f
y
x
f
xx
f
x
y
f
xy
x
f
Then, df = L+++++ 222
0
2
1
22
2
1
2 dyxdxdyydxdyxxydx
= L+++++ 222
)1(0
2
1
)1)(1)(1(2)1)(3(2
2
1
)1()1()1)(3)(1(2
...
2
1 2
2
2
+D+D= y
dy
Pd
y
dy
dP
dP
( ) L+
÷
÷
ø
ö
ç
ç
è
æ
¶
¶
+
¶¶
¶
+
¶
¶
+
¶
¶
+
¶
¶
=-++= 2
2
22
2
2
2
2
!2
1
,(), dy
y
f
dxdy
yx
f
dx
x
f
dy
y
f
dx
x
f
yxfdyydxxfdf

Bond Fundamentals
20
expectations that this rate will remain constant over the time interval of interest, you can use Taylor Series
to project the future value. Also, notice how adding more terms improves our estimate of the unknown
value. (Of course, the more information we have, the better). This is the theory underlying the convexity of
a bond idea, coming up in the next sections.

Bond Fundamentals
21
Bond Price Derivatives
Let’s delve deeper to see more precisely how price changes with yield. To do so, we have to take a
derivative. To make it easy on ourselves we will just use the constant cash flow C as from a coupon-paying
bond. Then,
If we factor out
y+
-
1
1
then the denominator will look like it did for P:
Now divide both sides by P. We then have
(The negative sign is used since we define positive duration as occurring when an increase in yield causes a
decrease in price, the normal result. We will see later that certain special fixed income instruments can
have negative duration.)
Dollar duration
Duration is used to make an estimate of how our bond’s price will change in response to a change in yield.
Duration measures the bond’s first-order sensitivity to a change in yield. It can most easily be thought of
as the change in price for a 100bp yield. The units of duration are time and it will have the same units as
the coupon payment interval (one year, one-half year, one-quarter year etc.)
Before moving to an example, a definition: “basis point”, or “bp” for short, is just an alias for “1/100 of a
percent.” Basis points are a frequent unit of measure in fixed income. There are 100bp per 1%, so
“100bp” is a way of saying “1%”.
( ) ( ) ( ) ( ) ( ) 11432
111
3
1
2
1 ++
+
-
+
+
-
++
+
-
+
+
-
+
+
-
=
¶
¶
nn
y
nPar
y
nC
y
C
y
C
y
C
y
P
L
( ) ( ) ( ) ( ) ( ) ( ) ( ) ú
ú
û
ù
ê
ê
ë
é
+
+
++
-=
ú
ú
û
ù
ê
ê
ë
é
+
+
+
++
+
+
+
+
++
-=
¶
¶
å
=
n
n
i
inn
y
nPar
y
iC
yy
nPar
y
nC
y
C
y
C
y
C
yy
P
111
1
111
3
1
2
11
1
1
32
L
( ) ( ) ( ) ( ) MACn
n
i
i
D
yy
nPar
y
iC
Pyy
P
Py
P
P +
-=
ú
ú
û
ù
ê
ê
ë
é
+
+
++
-=
D
D
@
¶
¶
å
= 1
1
11
1
1
111
1
Where
( ) ( ) ú
ú
û
ù
ê
ê
ë
é
+
+
+
= å
=
n
n
i
iMAC
y
nPar
y
iC
P
D
11
1
1
is defined as Macauley Duration.
The Modified Duration, DMOD, is defined as
( ) MACMOD D
y
D
+
=
1
1
, so MODD
y
P
P
-=
¶
¶1
( ) ( ) ( ) ( ) ( ) ( ) ( ) ú
ú
û
ù
ê
ê
ë
é
+
+
+
++
+
+
+
+
+¶
¶
=
+
+
+¶
¶
=
¶
¶
å
=
nnn
n
i
i
y
Par
y
C
y
C
y
C
y
C
yy
Par
y
C
yy
P
1111111 32
1
L

Bond Fundamentals
22
Example: If the Macauley duration of a bond is known to be 7.25 years when the price is $1,000, yield is
8% and the yield changes by 100 bp, what will be the change in the bond’s price? Just use the formula and
solve for DP:
before, this is the ten-year, 8% par bond. Since it is a par bond its price is known: $1,000. If the yield
So, the actual change in price is $1,000 – $935.82 = $64.18. (Notice how handy our little shortcut is.)
Here’s how to do this in Excel: We make a column of coupon payment times (i = 1,…, 10, for this
example); a column for cash flows (coupon rate/number of payments/year times par value); a column for
present value of each coupon payment (PVCF); and a column with time-weighted values of the cash flows,
i*PVCF. Then duration = the sum of i*PVCF over the sum of PV of cash flows.
Coupon Rate 8%
Par Value of Bond $1,000
Term (years) 10
Initial Yield 8%
Number of coupons/year 1
Coup Time Cash Flow PV of CF t * PVCF
1 $80 74.074074 74.07407
2 $80 68.587106 137.1742
3 $80 63.506579 190.5197
4 $80 58.802388 235.2096
5 $80 54.446656 272.2333
6 $80 50.41357 302.4814
7 $80 46.679232 326.7546
8 $80 43.221511 345.7721
9 $80 40.019917 360.1793
10 $1,080 500.24897 5002.49
Sum 1,000.00$ 7246.888
Macauley Duration 7.2469
Modified Duration 6.7101
Duration of a Bond
changes by 100 bp so that it is now 8% + 1% = 9%, the price of the bond will be
( ) ( )10
10
1 09.01
000,1$
09.01
80
+
+
+
= å
=i
i
P =
( ) ( )1010
09.01
1000
09.01
1
1
09.0
80
+
+
÷
÷
ø
ö
ç
ç
è
æ
+
- =$935.82.
MACD
y
yP
P
+
D
-=D
1
So, 25.7
08.01
)01.0(000,1$
+
-=DP =-$67.13. Let’s see how good a job this did. We already used this bond

Bond Fundamentals
23
Duration of zero coupon bond
For a zero coupon bond, the duration will be the same as the tenor of the bond, because we only receive one
cash flow and it’s at the end of the period. How sensitive are zeros to price changes? Since there are no
coupons, we can go back to basics and find that since
For a zero at par and yield of 8%, a 100 bp change in yield would cause a price change of
=$92.59. Because this is the same formula we had for the coupon-paying bond (except DMAC is replaced by
n, which is larger than DMAC for a coupon bond), the prices of zero coupon bonds are extremely sensitive to
changes in yield.
These durations are called dollar durations because they are expressed in terms of currency.
Dollar convexity
Now we need to talk about why there is an error between the change in price calculated using duration and
the actual change in price that would occur. If we plot bond price as a function of yield (again using our
10-year 8% bond) we get a graph like the following.
Bond Price as a Function of Yield
$0
$500
$1,000
$1,500
$2,000
0% 2% 4% 6% 8% 10% 12% 14% 16% 18% 20%
Yield, %
Price
( )nzero
y
Par
P
+
=
1
( ) ( )
nDn
yy
P
P
P
y
n
y
Par
y
n
y
nPar
y
P
MACnn
=Þ÷÷
ø
ö
çç
è
æ
+
-=
¶
¶
+
-=
++
-=
+
-
=
¶
¶
+ 1
11
,
1
1
11
1
1 1
08.01
)01.0(000,1$
10
1 +
-=
+
D
-=D
y
yP
nP

Bond Fundamentals
24
Note that the graph is not linear, but has a slight curve to it. This curve is known as convexity. This means
that as yields increase, the curve flattens: the bond price becomes less sensitive to changes in yield. When
yields are low, the price of the bond is extremely sensitive to changes in yield. Just using duration alone
assumes that the bond is equally sensitive to yield changes at any yield. So, we see that using duration
alone to estimate price sensitivity is not such a problem at high yields, but can lead to large errors when
yields are low. Using Duration alone to estimate price changes is reasonable only for small changes in
yield, where the price-yield curve can be assumed to be approximately linear.
Computing the price approximation
How do we include the effects of convexity in our price calculations? Recall the Taylor Series expansion.
We can expand price in terms of y in order to solve for DP. The expansion of P in terms of y is:
The second term is the adjustment that needs to be added to our price to account for the effects of
Because here the convexity is positive, convexity has value: it increases the price of the bond. Units of
convexity are in the unit of time, squared. A formula for use in a spreadsheet can be determined by taking
the second derivative of the P(y) equation with respect to y; the result is:
For example, re-computing the price change of our 10-year bond including the convexity of 60,531 gives
the total price change due to a change of 100bp in the yield as change due to duration + change due to
convexity:
( ) ( )2
01.0531.60
2
1
01.07101.6000,1$ +-=DP =-67.101+3.027=-$64.074. This is much closer to the actual
price change of P(0.09) – P(0.08) = 935.82-$1,000 = -$64.18.
Price value of a basis point (PVBP)
A common measure of duration is the price value of a basis point. You may sometimes see this referred
to as the dollar value of a basis point. This is a measure of bond price volatility. As the name implies,
( ) ( ) 2
2
2
2
1
y
y
P
y
y
P
yPyyP D
¶
¶
+D
¶
¶
+@D+ . Solving for DP and substituting our definition of duration gives:
( ) ( ) ErrorTruncationy
y
P
yPDy
y
P
y
y
P
PyPyyP MOD +D
¶
¶
+D-=D
¶
¶
+D
¶
¶
=D=-D+ 2
2
2
2
2
2
2
1
2
1
convexity. Defining dollar convexity as 2
2
y
P
C
¶
¶
= then ErrorTruncationyCyPDP MOD +D+D-=D 2
2
1
( )
( )
( )
( ) 2
1
22
2
1
1
1
1
+
=
+
+
+
+
+
+
=
¶
¶
å n
n
i
i
y
Parnn
y
Cii
y
P

Bond Fundamentals
25
this has to do with the price change resulting from a one-basis point, or 0.01%, change in yield. (We
calculated the price impact on our 10-year 8% bond resulting from a 100 bp change. This is the same
calculation but with a change of 1bp.)
Price of Bond at 8% Price of Bond at 8.01% Difference (PVBP)
$1,000 $999.33 0.67
Note that it does not matter if we increase or decrease the yield by 1bp; it is such a small amount that it
makes no difference. You should get the same answer either way. Sometimes this may be quoted “per
$100 of par value” so be aware of the conventions being used.
Estimating effective duration and effective convexity
If we divide the formula for DP by P, we get an expression for the percentage price change of the bond.
Then
where DE and CE are known as the effective duration and effective convexity of the bond, respectively.
(This duration is the same duration we have been using, except the convexity is now divided by P.)
If we have prices, we can estimate duration and convexity using finite-difference approximations of the
Example: A summary of three bond prices is summarized below.
Yield, % Price
7 1,070.24
8 1,000
9 935.82
The estimate of duration for a 100bp change in yield for our $1,000 bond at a yield of 8% is
2
2
1
yCyD
P
P
EE D+D-=
D
derivatives 2
2
,
y
P
y
P
¶
¶
¶
¶
. From Taylor Series Approximations these are derived as:
y
yyPyyP
y
P
D
D--D+
@
¶
¶
2
)()(
, so
yP
yyPyyP
DMOD
D
D--D+
-@
2
)()(
22
2
)()(2)(
y
yyPyPyyP
y
P
C
D
D-+-D+
@
¶
¶
= , 22
2
)()(2)(1
yP
yyPyPyyP
y
P
P
CE
D
D-+-D+
@
¶
¶
=
72.6
)01.0)(1000(2
24.107082.935
)01.0)(1000(2
%)1%8(%)1%8(
=
-
=
--+
-@
PP
DMOD

Bond Fundamentals
26
which is a pretty good approximation. Now, the convexity is calculated as:
This is also reasonably close to the 60,531 we calculated in Excel. The effective convexity is C/P = 60.6.
The percentage price change is then
so DP = -0.0642($1,000) = -$64.17 as before.
Portfolio duration and convexity
Suppose a fixed income manager is trying to decide between two portfolios. Portfolio A is a bullet
portfolio (one made up of bonds with maturities clustered at a single point on the yield curve) consisting of
our 10-year, 8% coupon bond. Portfolio B is a barbell portfolio (one made up of bonds with maturities
concentrated at both the short and long ends of the yield curve) consisting of a 2-year, 5% bond and a 20-
year, 12% bond. The manager is concerned with the effect of yield curve shifts on the performance of the
portfolios and wants to choose the best one. Since the duration of a portfolio of bonds is just the sum of
weighted durations of each bond (where the weights are the percentage held of each bond), it is easy to
choose the weights of portfolio B so that the duration matches the duration of Portfolio A. This is called a
duration-matched portfolio. We have the following durations:
Bond Mod Duration,
years
2 Year 1.86
10 Year 6.71
20 Year 7.47
For the barbell portfolio “B” we have DB = w2D2 + w20D20 = w2D2 + (1-w2)D20 = w21.86 + (1-w2)7.47. This
should be set equal to the duration of portfolio “A”, or 6.71 years. Solving for w2, we find w2 = 0.1355, so
13.55% is invested in the 2-year bond and 86.45% is invested in the 20-year bond. Now the fixed income
manager constructs scenarios of expectations of future yield curve shifts and wishes to know the price
change of the portfolios under each scenario.
Yield Curve Shift Scenarios
Yield Curve
Point
Scenario 1
shift, %
Scenario 2
shift, %
2 Year 100 100
10 Year 100 0
20 Year 100 -100
600,60
01.0
24.1070)000,1(282.935
01.0
%)7(%)8(2%)9()()(2)(
222
=
+-
=
+-
=
D
D-+-D+
@
PPP
y
yyPyPyyP
C
0642.0)01.0)(6.60(5.0)01.0(72.6
2
1 22
-=+-=D+D-=
D
yCyD
P
P
EE

Bond Fundamentals
27
So portfolio B will outperform Portfolio A if this scenario occurs. Note that portfolios of equal duration
will experience the same price change (neglecting convexity effects) only when the yield curve undergoes a
parallel shift. If the yield curve shifts in a non-parallel manner, the price change will no longer be the
same. This illustrates the fact that duration is just a measure of interest rate risk (the risk that the yield
curve will shift in a parallel manner), and not yield curve risk (the risk that the yield curve will shift in a
non-parallel manner.) Including convexity will not solve this problem. A good way to deal with yield
curve risk is to construct the portfolio using key rate durations (also known as “partial durations”). Key
rate duration is portfolio duration calculated using certain so-called “key rates.” Key rate duration is a
measure of the sensitivity of a bond to a single point on the yield curve. For example, if the sensitivity of a
5-year Treasury to a 100-bp change in the yield curve is desired, you would shift the zero curve by 100 bp
and calculate the new price of the bond. The key rate duration is given by the percentage change in the
price of the bond. Specific key rates may be found, for example, on Bloomberg.com. Key rate duration is
often referred to as partial duration because, in calculating it, you only consider exposure to a section of the
yield curve. Key rate durations can be used to assess the effect of any yield curve shifts on the portfolio.
Scenario 1 For Bond A, we have )01.0(71.6-=D-=
D
yD
P
P
EA =-6.71%. For Bond B,
yDwyDwyDw
P
P n
i Eii D-D-=D-=
D
å = 2020221
=-0.1355(1.86)(0.01)-0.8645(7.47)0.01 = -6.71%. It
should not come as a surprise that the price change in both cases is the same, since this is why we
constructed portfolio B with the weights that we did.
Scenario 2 For Bond A, there is no change in price since the 10 year point on the yield curve does not
shift. For Bond B, we have
)01.0)(47.7(8645.001.0)86.1(1355.0202022 ---=D-D-=
D
yDwyDw
P
P
=6.21%.
We have that 2
2
1
yCyD
P
P
EE D+D-=
D
. We will ignore convexity changes in the following, but they
could very easily be included.

Bond Fundamentals
28
Sample Questions and Answers
Questions
1. What is the duration of a 10-year zero coupon bond?
2. Which is more volatile, a 30-year zero coupon bond or a 30-year 6.5% coupon bond?
3. What’s the value of a stock if it currently pays a dividend of $2.50, the hurdle rate is 15% and the
dividend is growing at an expected rate of 3%/year?
4. If you saw a 5-year, 10% coupon bond yielding 10% listed for $995, would you buy it? Assume
that the coupons are paid annually and use simple compounding.
5. What happens to bond prices when interest rates increase? Why?
6. If you are a trader and have an idea that rates will decrease, what strategy could you adopt to profit
from this?
7. Which should be cheaper and why: a 20-year, 7% coupon bond, or a 20-year, 8% coupon bond?
8. If you have a bond and it is $100 at a yield of 6%, $95 at 7%, and $90 at 8%, what is the modified
duration? What is the effective convexity?
9. What is the duration of a fixed income ladder portfolio consisting of a 5-year bond with a duration
of 4.6, a 10-year bond with duration of 7.2, and a 20-year bond with duration of 14.3? Assume
that you hold equal proportions of all bonds.
10. What would be the impact of a 100 bp increase to the yield curve to a par bond with duration of
7.52? Is this the price you would see on a Bloomberg? Why or why not?
11. What is duration? Is it constant for all yields? What is convexity and why is it important?
Answers
1. 10 years. No calculations required. The duration of an n-year zero is just n.
2. The zero coupon bond is the most volatile.
3. Use the Gordon Growth model: PV = CF0/(r-g) = $2.5/(.15-.03) = $20.83.
4. Yes, because this is a par bond (coupon equals yield) and should be trading at par ($1,000).
5. Except for rare exceptions (some types of mortgage bonds, for example) bond prices decrease
when interest rates increase. If you had a 5-year par bond paying 8% coupons, and suddenly the
5-year treasury went to 9%, your bond would be worth less than before because investors would
be getting less than the current rate.
6. If you think rates will drop, this implies that bond prices will increase. You should buy bonds
(zero coupon in particular.)
7. You don’t really need to do any calculations since the maturities and (we assume) the rest of the
variables are the same for each bond, except for the coupon. The 7% coupon bond should be
cheaper since, ceteris paribus,
( ) ( ) ( ) ( )N
n
i iN
n
i i
rrrr +
+
+
<
+
+
+
åå ==
1
100
1
8$
1
100
1
7$
11
8. A straightforward calculation: use the definition (see section “Estimating Effective Duration and
Convexity”). Make a table:

Bond Fundamentals
29
Yield, % Price
6 100
7 95
8 90
(Is this surprising? The price appears to be linear with yield: each 1% change in yield results in a $5
change in price, at least in this region. Had this been noticed, no calculations would have been
necessary.)
9. Straight-forward calculation: DP = w5D5 + w10D10+w20D20 = 1/3(D5 + D10 + D20) =
1/3(4.6+7.2+14.3)=8.7
10. Interviewer is looking for a straight definition, and interpretation:
The new price would be calculated as $1,000 - 75.2 = $924.8. But this is not the price that would be
shown on Bloomberg because we have neglected convexity and higher-order terms. Duration is just a
first-order approximation.
11. Duration is a measure of the bond’s sensitivity to yield curve movements. It is defined as:
It is steep when yields are low, and flattens as yields get higher, meaning the sensitivity of a bond’s
price to changes in yields decreases as yields increase. The price-yield curve is convex and the
duration gives a good approximation only for small changes in yield where the price-yield curve can be
assumed to be close to linear. Large errors occur when Dy increases. To account for the curvature of
price with respect to yield, we need to include a second-order term. This is known as convexity, the
second derivative of price with respect to yield. Since the second derivative is positive over the curve,
convexity increases price. If it is neglected, the price calculated using duration alone would be too
low.
yP
yyPyyP
DMOD
D
D--D+
-=
2
)()(
. Since we are given data +/- 1% from a center point of 7%, Dy has
to be 1%, there is no other choice. Then
)01.0)(2(95
10090
)01.0(2
%)1%7(%)1%7(
%7
-
-=
--+
-=
P
PP
DMOD =5.26
years. For convexity,
2
)()(2)(1
y
yyPyPyyP
P
CE
D
D-+-D+
=
( )22
01.0
100)95(290
95
1
)01.0(
%)6(%)7(2%)8(
95
1 +-
=
+-
=
PPP
=0.
MODMOD yDPPso
y
P
Pdy
dP
P
D D-=D
D
D
-@-=
11
=-$1,000(0.01)7.52 = -$75.2.
dy
dP
P
DMOD
1
-=

Bond Fundamentals
30
Summary of Formulas
Macauley Duration
( ) ( ) ú
ú
û
ù
ê
ê
ë
é
+
+
+
= å
=
n
n
i
iMAC
y
nPar
y
iC
P
D
11
1
1
Modified Duration
( ) MACMOD D
y
D
+
=
1
1
Price Change of a Bond MODD
y
P
P
-=
¶
¶1
Price of a Bond
( ) ( )nn
r
Par
rr
C
P
+
+
÷
÷
ø
ö
ç
ç
è
æ
+
-=
11
1
1
Gordon Growth Model (Perpetuity with growth rate g)
gr
CF
PV
-
= 0
Present Value of Perpetuity
r
CF
PVPerpetuity =
Future Value of P0
nt
t
n
r
PP ÷
ø
ö
ç
è
æ
+= 10
using continuous compounding: rt
t ePP 0=
Present Value of Pt
nt
t
n
r
PP
-
÷
ø
ö
ç
è
æ
+= 10
Using continuous compounding: rt
t ePP -
=0
Present Value of Stream of Cash Flows
( )
å =
+
=
n
i i
i
i
r
CF
PV 1
1
PV of Annuity (short cut)
( ) r
C
rr
C
PV n
+
-=
1
1
Dollar Convexity 2
2
y
P
C
¶
¶
=

Bond Fundamentals
31
Numerical Approximations of Duration and Convexity:
Price Change of Bond due to Change in Yield ErrorTruncationyCyPDP MOD +D+D-=D 2
2
1
Duration
yP
yyPyyP
DMOD
D
D--D+
-@
2
)()(
Convexity 22
2
)()(2)(
y
yyPyPyyP
y
P
C
D
D-+-D+
@
¶
¶
=
Taylor Series Expansion (Single Variable)
( ) ( ) M
MM
n
n
n
Rkxkf
M
kxkfkxkfkxkfkfkx
n
kf
xf +-++-+-+-+=-= å
¥
=
)(
!
1
))(('''
!3
1
))((''
!2
1
))((')(
!
)(
)( )(32
0
)(
L

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STATISTICS

Statistics
35
Random Variables
Statistics concerns itself with distributions and properties of random variables. A random variable can be
thought of as a drawing from a distribution whose outcome prior to the draw is uncertain, or stochastic.
The outcome of a roll of a pair of dice, the next song that will be played on the radio, what the Fed will do
in next Tuesday’s Open Market Committee meeting, next month’s peso/USD exchange rate, and next
year’s return on the S&P500 index are all random variables, because the outcome is uncertain prior to the
occurrence of the event. Although, of course, we cannot predict the outcome of random variables, we can
form expectations of them. And forming expectations of random financial variables is at the very heart of
finance. We will be developing the properties of these variables as we move through this chapter. First,
we need to distinguish between discrete and continuous random variables.
A discrete random variable can take on a countable number of values (that is, finite). Discrete random
variables include the roll of dice in a craps game, roulette outcomes and lottery picks and. For a roll of a
single die, the ultimate outcome is unknown prior to the event taking place, but we know that there are six
possible outcomes: either 1, 2, 3, 4, 5 or 6 will be rolled. There are no other possibilities. We have
covered them all. One and only one of these outcomes will ensue. These facts are required in all of the
distribution of outcomes. The set of possible outcomes is called the event space, or states of nature.
The set of outcomes must be mutually exclusive and collectively exhaustive. When we assign a
probability to each possible state of nature, we can then form expectations of the random variable.
A continuous random variable can take on any value in a given range, which may be (-∞, +∞). Examples
of continuous outcomes would be those variables that have an uncountable (or infinite) number of possible
outcomes. These would include stock returns. You could imagine a return of –10%, -9.9%, -9.99%, -
9.999%, …, 1.45%, 46%, … and any value, actually. With leverage you could even go beyond –100%.
Tomorrow’s temperature is a random variable that will be drawn from a random distribution of possible
temperatures. When senior manager Gordon ponders the future value of his Lucent stock options, he might
use a continuous distribution.
Rules governing probabilities associated with distributions
1. All probabilities must be greater or equal to zero and less than or equal to one.
2. The sum of all probabilities (discrete distribution) or integral of probabilities (continuous
distribution) must equal one.
3. The set of outcomes must be mutually exclusive and collectively exhaustive.
No matter the form of the random variable, we need to think about what type of distribution the outcomes
will have. For the discrete variable, hopefully we can enumerate all outcomes and assign expected
probabilities to them. For example, assuming we have a fair die, there is an equal probability of each
outcome {1,2,3,4,5,6} occurring. Since there are six possible outcomes, there is a one in six probability of
each outcome. It gets a little more complicated when we talk about a continuous random variable. We
might do some scenario analysis and estimate that the probability of losing 10% on our portfolio is 23.1%.
But it is unlikely that the ultimate return will be exactly what we expect. Instead, we can calculate the
probability that the return will fall in some range about the expected outcome. To do this, we need to have
a good idea of the nature of the distribution of returns.
Univariate distribution functions
Each random variable is a draw from some distribution, even if the distribution is unknown. The draws can
be “without replacement” or “with replacement.” The roll of a die or a return on a stock is sampled with
replacement, while a lottery drawing is preformed without replacement. We also expect successive draws

Statistics
36
to be “independent.” The flip of a coin, the roll of a die or a return on a stock should not depend on what
happened before. Distributions of outcomes of independent random variables should exhibit no pattern or
memory.
There are other distributions where successive outcomes depend on what happened before, such as random
walks and Markov chains. In statistics, you often consider processes that have no memory: the value of
the next (unforeseeable) observation is independent of past history; it depends only on the current level of
the variable. So, if stock prices are Markovian, this means that prior history of the stock’s price process
does not help predict future values. The best estimator of the next observation is the current observation.
Random walks are examples of Markov chains, as are outcomes from roulette wheels, craps and other (fair)
games of chance. Some lottery scheme vendors claim that future lottery picks can be guaranteed by
studying the past numbers drawn. They look for patterns in the data, in a way similar to technical traders.
Discrete density function
If we let the symbol X represent the random variable “outcome of a roll of dice,” then the event set of the
random variable X is the set x ={1,2,3,4,5,6}. Each outcome of this set is equally probable. There are six
total outcomes, so the probability distribution associated with the outcome set is {1/6,1/6,1/6,1/6,1/6,1/6}.
A univariate distribution function is a distribution function associated with a single random variable,
such as this example of the roll of a die. Another term for this type of probability distribution is discrete
density function. This particular example is called a uniform density since all outcomes are equally
probable: if we graphed them, we would just have a horizontal line. Note that the area under this curve is
one, as required: area = base*height = 6*1/6 = 1.
Another common example is the tossing of a fair coin. Each toss can have one of two outcomes, either tails
(T) or heads (H). Like the roll of a die, a successive toss is independent of the prior result. Each outcome
has a probability of 1/2 of occurring.
Multivariable distribution functions
It is very interesting to think about what happens if we combine two (or more) independent random
variables. What will the resulting probability distribution and outcome space look like? The resulting
distribution of outcomes of such experiments is called a joint probability distribution. A familiar
example is the game of craps, in which two die are rolled simultaneously. Each die has its own distribution
of outcomes and probabilities. We already found that the set of outcomes is a uniform distribution when a
single die is rolled. But what happens when we roll two simultaneously? First of all, notice that we can’t
get a result of “1” anymore. However, we can get a “12,” which was impossible before. The possible
outcomes in this case are the sum of each individual outcome,
But does the distribution still look uniform? That is, does each outcome {2,3,4,5,6,7,8,9,10,11,12} have
equal probability of occurring? The answer to this question is crucial to gambling success.
To begin to figure this out, I make a table of all of the possible outcomes:
Die 1 is …
Die 2 is …
1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
åå
= =
+=
6
1
6
1i j
jix

Statistics
37
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
Now we tally up the results.
Result Number of
Ways to
Occur
Set of Outcomes Leading to Result Probability of
Occurring
2 1 {1,1} 1/36
3 2 (1,2},{2,1} 2/36
4 3 {1,3},{2,2},{3,1} 3/36
5 4 {1,4},{2,3},{3,2},{4,1} 4/36
6 5 {1,5},{2,4},{3,3},{4,2},{5,1} 5/36
7 6 {1,6},{2,5},{3,4},{4,3},{5,2},{6,1} 6/36
8 5 {2,6},{3,5},{4,4},{5,3},{6,2} 5/36
9 4 {3,6},{4,5},{5,4},{6,3} 4/36
10 3 {4,6},{5,5},{6,4} 3/36
11 2 {5,6},{6,5} 2/36
12 1 {6,6} 1/36
(Notice that the strategies “draw a figure,” “look for a pattern,” “enumerate all cases,” and even “exploit
symmetry” could have been used to solve this.) What are the probabilities of each outcome? To find out,
we first sum up the number of possible outcomes: this is the sum of numbers in the second column, or 36.
The probability of each individual result occurring is just the number of ways each can occur, divided by
the total number of outcomes -- this is a general result. And you will see that if you check this, the
probabilities sum to one. The probabilities for our game are shown in the last column above. The
probability of “1/36” for rolling a 12, for example, means that there is only one way to roll a 12, out of 36
possible outcomes. So the next time you see someone buying a 12 in Vegas, you will know that this is a
low probability bet. The number having the highest frequency (probability of occurring) is 7. You can see
that the probability distribution is definitely not uniform. If you graph it, it will look like the following:
Probability Distribution of Outcomes in Craps
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
2 4 6 8 10 12
Outcome of Roll of Two Die
ProbabilityofOutcome

Statistics
38
The distribution does not appear to be linear, so we conclude that the outcome of the sum of two uniform
distributions is not uniform. Even though this may seem like a simple example, there is a lot that we can
learn from it -- just because it is so simple, and we can completely describe and understand it.

Statistics
39
Key Statistical Figures
It is often desired to summarize large sets of data so that they may be better understood and compared. To
do this, we want some kind of measure of central tendency of the data. There are three measures of
central tendency that will be of interest: the mean, the median and the mode. These measures are
important, because we will use them to help us determine the expected value of a random variable. We will
also use simple descriptive statistics, including range, minimum, maximum, interquartile range,
variance, standard deviation and skewness.
In the following we consider a time series of the monthly returns of a particular stock over the past nine
months {8%,-10%,0%,5% ,-6%,-6%,7%,13%,6.5%}. It’s always a good idea to graph the data you are
trying to understand to look for tendencies and trends, and to visually identify any outliers.
The data appear to be clumped toward the right of the x-axis. The following is a description of the
measures of central tendency of the data:
Mean
The mean of a set of numbers is the same as its average. The mathematical definition of mean is
We use an overbar on x to represent the mean of all of the xi. In statistics, when speaking of distributions,
the symbol m is usually used. The mean of a distribution is very important. In many cases, this is the best
estimate we have of an expected future value.
100-10
C2
Dot Plot of Stock Returns (%)
n
x
x
n
i
iå
=
= 1

Statistics
40
Example What is the mean of our stock market returns?
Just sum up all of the values and divide by the number of data points, or 0.175/9=1.94%.
Example What is the mean of the outcomes of the roll of two simultaneous die?
Sum up 2,3,4,5,6,7,8,9,10,11 and 12 and divide by 11 to get 7.
(Quick Trick: to add a sequence of numbers increasing by one and running from 1 to n, the formula is
( )
2
1+nn
. For example, the sum of integers 1, 2, 3, …, 99,100 is (100)(101)/2 =5050. To do our problem,
we have 2, 3,4, …, 11,12, which almost matches the required form but is offset by one. Just shift back by 1
to do the sum 1, 2, … , 11 = 66, average = 6, then add back 1 to get our average, 7.) Is it a coincidence that
the mean result is also the most frequent result? If you roll a pair of die, the combination 7 is more likely to
occur than any other combination. But this is also the most frequent observation. It turns out that, in this
case, the mean is the same as the median and the mode.
Median
To calculate the median, the data must first be sorted in ascending order. The median is the middle value
of the sorted series. The median is useful because half of the data will lie below it and half above it.
Comparison of the median to the mean gives us an idea of the skewness of the distribution. The sorted data
are {-10%,-6%,-6%, 0%, 5%, 6.5%, 7%, 8%, 13%}. We have nine data points, so the median is the fifth
number in the sorted series of numbers, or 5%. This means that half of the returns lie below 5% and half
lie above 5%. So, if I wanted to know the expected return of the stock next month, which number should I
use? The mean of 1.94%? The median of 5%? The most recent return of 6.5%? How about the return
occurring most frequently?
Mode
The mode tells us which response occurs with greatest frequency. Distributions can have a single mode, no
mode or multiple modes. In the roll of a single die, all responses occur with the same frequency, so there is
no mode. In the roll of two dice, the outcome “7” occurs with greatest frequency. In the stock return
example, we have a return of –6% occurring as the mode.
Range, maximum and minimum
We would now like an idea of the dispersion of the data. The range provides a simple measure of the
dispersion of the data and is defined as
Maximum – Minimum values of data set.
The maximum and minimum of a data set are interesting numbers in themselves. For the return data, the
maximum is 13%; the minimum is –10%. This makes the range 23%. The range of the roll of a pair of die
is 12-2=10. The range provides an idea of how tightly clustered the data are about the mean, particularly
when it is normalized by dividing by the sample standard deviation (which we’ll discuss later in this
chapter). (There was a question about the range of data on this year’s CFA level II exam that tripped up
many people who were not expecting to see it.) However, the range will be skewed by excessively large or
small data elements. To do a better job we can use the interquartile range.

Statistics
41
Interquartile range
The data can be sorted and divided into equally sized groups. If there are outliers, they can be removed if
desired, and the remaining data classified. If n sorted data points are to be divided into four groups, the
first quartile Q1 is the data point given by
4
1+n
. The third quartile Q3 is the th
n
4
)1(3 +
point, and the
median Q2 is the th
n
4
)1(2 +
point. If the number we calculate for the Qi isn’t an integer, interpolation is
used. The first quartile for the stock return data is 10/4 th
point = 2.5th
point. This means to go halfway
between the second and third point. Since both are equal to –6, Q1 = -6. Q3 = 3*2.5 = 7.5. Again we
interpolate midway between the seventh and eighth points to get Q3 = (7+8)/2 = 7.5. The interquartile
range is defined as the difference Q3 – Q1, here 7.5 – (6) = 13.5. This is often thought of as a better
measure of dispersion than the range, because excessive values that may lie on the extremes of the data set
(such as may occur in distributions of wealth) will not skew the result as they are excluded from the
interquartile range.
Variance
The best measure of dispersion from the mean is the variance ( s2
). This is defined as the sum of squared
differences of each data point from the mean. These differences are called deviations from the mean.
Then
Notice that we divide by n-1, not n, in computing the variance. There are deep statistical reasons for this
(see the following paragraph**), but the main reason is that we really don’t know the true population
(except in simple, constructed cases such as craps or flips of coins). We are just estimating the variance of
the unknown, true population with the variance of our sample. It turns out that this means we have to
divide by n-1.
Month i Ri (%) Ri - m (Ri - m)
2
1 -10 -11.94 142.67
2 -6 -7.94 63.11
3 -6 -7.94 63.11
4 0 -1.94 3.78
5 5 3.06 9.34
6 6.5 4.56 20.75
7 7 5.06 25.56
8 8 6.06 36.67
9 13 11.06 122.23
Sum 17.5 487.22
Mean m 1.944
Variance 60.90
( )
11
1
2
1
2
2
-
=
-
-
=
åå
==
nn
yy
n
i
i
n
i
i d
s

Statistics
42
**Suppose that we actually have the true population. Then, the variance would be computed as
( )
n
x
x
i
iå
=
-
= 1
2
2
m
s , where m = population mean and n is the total number of points. However, in practice
we do not know the mean m, but can only approximate it with the mean of the sample we take, x . This
sample mean may or may not be reflective of the true population mean, so we divide by n-1 rather than n to
compensate for this fact. This way, the sample variance
( )
1
1
2
2
-
-
=
å
=
n
xx
s
x
i
i
that we compute will hopefully
be, neither larger nor smaller than the true population variance s2
.
Standard deviation
The standard deviation (s) is the true measure of dispersion. It is just the square root of the variance, equal
to 7.804% in the preceding example. (In Excel: =STDEV(range of numbers)). For the roll of a single
die, the standard deviation is 1.87. We will find that normal distributions can be uniquely described by the
mean and the standard deviation. The standard deviation is used as a measure of risk in finance. If
portfolios of assets are normally distributed, they can be ranked according to return and risk. This is the
basis of the efficient markets hypothesis. In a normal distribution, there is a 68.2% probability that the
data will lie within one standard deviation of the mean; a 95.5% probability that the data will lie within two
standard deviations of the mean, and a 99.7% change that the data will lie within three standard deviations
of the mean. For the roll of a die, m – s = 3.5-1.87 = 1.63 and m + s = 3.5+1.87 = 5.37. This means that
about two-thirds of the data, or four data points, should lie between 1.63 and 5.37 (or 2 and 5 in integer
values), which they do -- even though the data are not normally distributed.
(Quick trick: To remember whether variance or standard deviation is equal to s, just remember that s
starts with one s. So does standard deviation. Variance, then, must be s2
.)
Correlation
We have talked about simple, independent experiments, but financial data are often correlated. This means
that the data move somewhat together, influencing each other. Since the S&P500 is made up of 500 large-
cap stocks, factors that influence large caps will also move the index. Bond prices are sensitive to interest
rates. Automakers Ford and GM are probably influenced by common macroeconomic factors. Gold and
silver may have some association. Sometimes assets move in opposite directions, such as a call on a stock
and a put on the same stock, or holding gold in a portfolio that is heavy in risky tech stocks having high
market bs. The degree of association or comovement between two variables is called correlation (r).
Another term that can express this idea is covariance. The degree of covariance between assets within a
portfolio must be included when calculating the variance of a portfolio.
The correlation coefficient r can take on values from –1 to +1, with –1 meaning perfect negative correlation
and +1 meaning perfect positive correlation. A value of zero implies no correlation, while low values
indicate virtually no association between the variables.
The definition of correlation between two variables x and y is:
( )( )
( )
11,
1
1
££-
-
--
=
å
=
r
n
yyxx
r
yx
n
i
ss

Statistics
43
Perfect Positive Correlation
-30%
-20%
-10%
0%
10%
20%
30%
-30% -20% -10% 0% 10% 20% 30%
x
y
Perfect Negative Correlation
-30%
-20%
-10%
0%
10%
20%
30%
-30% -20% -10% 0% 10% 20% 30%
x
y
Virtually No Correlation
5%
5%
5%
6%
6%
6%
6%
-30% -20% -10% 0% 10% 20% 30%
x
y

Statistics
44
As an example, we calculate the correlation between monthly change in a hypothetical airline’s profit as a
function of change in fuel prices. The graph of the data is shown below.
Impact of Change in Fuel Prices on Airline Profit
-0.15
-0.1
-0.05
0
0.05
0.1
-15% -10% -5% 0% 5% 10% 15% 20% 25% 30%
Monthly Change in Fuel Prices, %
MonthlyChangeinProfit,%
x y
Month i
Change in
Price of
Airline
Fuel $/gal
Airline
Profit
Change
(%/qr) x-mx y-my (x-mx)(y-my)
1 -10% 8.3% -17.5% 11.8% -2.1%
2 -5% 1.9% -12.5% 5.4% -0.7%
3 0% 0.0% -7.5% 3.5% -0.3%
4 5% -1.8% -2.5% 1.6% 0.0%
5 10% -10.0% 2.5% -6.5% -0.2%
6 15% -6.9% 7.5% -3.4% -0.3%
7 20% -7.5% 12.5% -4.0% -0.5%
8 25% -11.9% 17.5% -8.5% -1.5%
Sum 60.0% -27.9% -5.4%
Mean 7.5% -3.5%
Stdev s 12.25% 6.82%
Correlation -0.93138588
There seems to be a negative correlation, as we would expect. Using the formula will enable us to assign a
quantitative value to this association.
( )( )
( ) %)82.6%)(25.12)(7(
%4.5
1
1 -
=
-
--
=
å
=
yx
n
i
n
yyxx
r
ss
=-0.9313

Career Guide Advanced Finance and Quant Finance Interviews

Career Guide Advanced Finance and Quant Finance Interviews

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